Question

Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R.

Prove that

1. AQ = QR
2. AP = 2PQ
3. PR = 2AP

Answer 1

Given: A parallelogram ABCD and Q is the mid-point of CD

i. In ΔAQD and ΔRQC

∠DQA = ∠CQR   ...[Vertically opposite angles]

∠DAQ = ∠CRQ   ...[AD || BC and alternate interior angles]

⇒ ΔAQD ∼ ΔRQC   ...[AA similarity]

⇒ `(AQ)/(QR) = (DQ)/(CQ)`

But CQ = DQ

⇒ AQ = QR   ...(1)

Hence, proved.

ii. In ΔAPB and ΔQPD

∠APB = ∠QPD   ...[Vertically opposite angles]

∠PQD = ∠PAB   ...[CD || AB and alternate interior angles]

⇒ ΔAPB ∼ ΔQPD

⇒ `(AP)/(PQ) = (AB)/(QD)`

But AB = CD = 2QD

⇒ `(AP)/(PQ) = (2QD)/(QD) = 2/1`

⇒ AP = 2PQ   ...(2)

Hence, proved.

iii. Since, PR = PQ + QR

PQ + AQ   ...(From 1)

PQ + AP + PQ = 2PQ + AP 

= AP + AP   ...(From 2)

= 2AP

Hence, proved.